Question

Short answer

The result of the speed of the second projectile is \({v_2} = 1.414{v_1}\).

Step-by-step solution

Initial projectile speed

Use the initial projectile speed relation and then compare the two speeds of projectiles with identical mass to calculate the speed of the second projectile.

Given data

Given data:

The maximum height of projectile 1 is\(h = 2.6\;{\rm{cm}}\).

The maximum height of the second projectile 1 is \({h_2} = 5.2\;{\rm{cm}}\).

Determine the fraction of speed

The formula of initial projectile speed is given by the following equation:

\(v = \frac{{m + M}}{m}\sqrt {2gh} \)

Here,m and M are the mass of the first and second projectile, respectively.

The fraction of projectile speed is given by the following:

\(\begin{array}{c}\frac{{{v_2}}}{{{v_1}}} = \frac{{\frac{{m + M}}{m}\sqrt {2g{h_2}} }}{{\frac{{m + M}}{m}\sqrt {2gh} }}\\\frac{{{v_2}}}{{{v_1}}} = \frac{{\sqrt {{h_2}} }}{{\sqrt h }}\end{array}\)

Determine the speed of the second projectile

On plugging the values in the above equation,

\(\begin{array}{l}\frac{{{v_2}}}{{{v_1}}} = \frac{{\sqrt {5.2\;{\rm{cm}}} }}{{\sqrt {2.6\;{\rm{cm}}} }}\\\frac{{{v_2}}}{{{v_1}}} = 1.414\\{v_2} = 1.414{v_1}\end{array}\).

Thus, \({v_2} = 1.414{v_1}\) is the speed of the second projectile.